sysLoader.getResource() problem in java

Question

I am having following lines of code.

sysLoader = (URLClassLoader)Thread.currentThread().getContextClassLoader();
url = sysLoader.getResource("tempFile.txt");

It is giving an weird problem. If I run this from a path where there is no space in the path (Folder names) then it is running fine. But if the path contains any spaces (line "c:\New Foler...") then it is not working.

How to solve this?

EDIT: In more detail - I inspected the sysloader object.

sysloader -> UCP -> path

Is having a path with character %20 instead of space

And therefore all the URLs are null.

How to resolve this?

Bow

Define "not working". Does it return null? Or does the returned URL not work in some context? – Gordie

Answer 1

This is known by Sun/Oracle, their advice is to use URI objects which will remove the %20 characters:

Instead of doing this:

FileInputStream fis = new FileInputStream(url.getFile());

you can force any %-escaped characters to be decoded by first converting the URL to a URI, and then use the path component of the URI as the filename:

URI uri = new URI(url.toString());
FileInputStream fis = new FileInputStream(uri.getPath());

Doyen

Here is the corresponding but report marked as "Won't fix": bugs.sun.com/bugdatabase/view_bug.do?bug_id=4466485 . This bug/feature still exists in Java 1.7 . – Risk

A shorter version without explicitly converting to string and back to URI would be: URI uri = url.toURI(); FileInputStream fis = new FileInputStream(uri.getPath()); – Ingeborg

Answer 2

Use URLDecoder.decode() method to replace %20 characters by spaces.

String path = URLDecoder.decode(url.getPath(), "UTF-8");

Please also keep in mind that when resource is located in jar file you have to handle it different way. See it e.g. here: How to access resources in jar where it can be present in multiple jar

Lorrin

Once you get the URL from getResource that is used, while passing the path in the file, we can use the method URLDecoder#decode. Something like this: File file = new File(URLDecoder.decode(testFile.getFile(), "UTF-8")); Here, testFile is a URL object. I am adding a comment since even though I found the answer useful, I had to somehow think about how to use this. Thanks anyway. – Bickerstaff

Answer 3

To get the URL of the file from string, when the path contains spaces, this is what worked for me:

File file = new File("/Users/work space/tempFile.txt");
URL url = file.toURI().toURL();

According to Javadocs, file.toURL() is deprecated:

This method does not automatically escape characters that are illegal in URLs. It is recommended that new code convert an abstract pathname into a URL by first converting it into a URI, via the toURI method, and then converting the URI into a URL via the URI.toURL method.

Hence used file.toURI().toURL().

For Java 7+, this is approach can be taken instead:

URL url = Paths.get("/Users/work space/tempFile.txt").toURI().toURL());

Note: If the path begins with a / it is considered absolute else taken as a relative path.

Wacker